What is the domain of y=sqrt(3x-1) ?

The domain of y=sqrt(3x-1) is x>= 1/3

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domain of y=sqrt(3x-1)

domain

y = 3 x - 1

x \geq \frac{1}{3}

Interval Notation

[\frac{1}{3}, \infty)

Solution steps

Find non-negative values for radicals:

x \geq \frac{1}{3}

The function domain

x \geq \frac{1}{3}

(x - 2) \cdot \frac{1 + x}{1 - x}

f (x) = - 2 x^{2} + 4 x + 6

f (x) = \frac{1}{\frac{x}{2} + 6}

\frac{2 x - 1}{x ^{2} - 5 x + 6}

f (x) = - 3 x^{2} + 360 x